<u>Answer:</u> The value of
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of
= 9.2 g
Molar mass of
= 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:

For the given chemical equation:

<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of
= 0.057
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 1.435
Answer:
Explanation:
The main task here is that there are some missing gaps in the above question that needs to be filled with the appropriate answers. So, we are just going to do rewrite the answer below as we indicate the missing gaps by underlining them and making them in bold format.
SO; In the quantum-mechanical model of the hydrogen atom.
As the n level increases. the energy <u>increases</u> and thus levels are <u>closer to </u>each other. Therefore, the transition <u>3p→2s</u> would have a greater energy difference than the transition from <u>4p→3p.</u>


Answer
5
Explanation:
We can go about this using the percentage compositions.
First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.
0.96/1.5 * 100 = 64%
Hence the percentage by mass of the water present is 36%
The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol
The molar mass of the water is 2(1) + 16 = 18g/mol
Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles
The total mass of the copper sulphate hydrate is 160+ 18x
Now how do we get x? Like it is said earlier, the percentage composition is constant.
Hence, 64/100 * (160 + 18x) = 160
16000 = 64(160 + 18x)
16000 = 10,240 + 1152x
16,000 - 10,240 = 1152x
1152x = 5760
x = 5760/1152
x = 5
C5H12 + 8 O2 → 5 CO2 + 6 H2O
8 molecules of O2 are required.
Empirical formula of a compound gives the proportions of the elements in that compound but it does not define the actual arrangement and number of atoms.
Let the empirical formula of compound be
.
The ratio of number of moles of C, H and O can be calculated as follows:

Simplifying the ratio,

Thus, the value of x, y and z will be 1, 2 and 1 respectively.
Therefore, the empirical formula will be
.